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Feedback Control System
Dr.-Ing. Erwin Sitompul
1/2Erwin Sitompul Feedback Control System
Textbook and Syllabus
Textbook:Gene F. Franklin, J. David Powell, Abbas
Emami-Naeini, “Feedback Control of Dynamic Systems”, 6th Edition, Pearson International Edition.
Syllabus:1. Introduction2. Dynamic Models3. Dynamic Response4. A First Analysis of Feedback5. The Root-Locus Design Method6. The Frequency-Response Design Method
IDR 192,000
USD 112.50
1/3Erwin Sitompul Feedback Control System
Grade Policy
Final Grade = 10% Homework + 20% Quizzes + 30% Midterm Exam + 40% Final Exam + Extra Points
Homeworks will be given in fairly regular basis. The average of homework grades contributes 10% of final grade. Homeworks are to be written on A4 papers, otherwise they
will not be graded. Homeworks must be submitted on time. If you submit late,
< 10 min. No penalty10 – 60 min. –20 points> 60 min. –40 points
There will be 3 quizzes. Only the best 2 will be counted. The average of quiz grades contributes 20% of final grade.
1/4Erwin Sitompul Feedback Control System
Midterm and final exam schedule will be announced in time. Make up of quizzes and exams will be held one week after
the schedule of the respective quizzes and exams, at the latest. The score of a make up quiz or exam can be multiplied by 0.9
(the maximum score for a make up is 90).
Extra points will be given every time you solve a problem in front of the class. You will earn 1 or 2 points.
Grade Policy
Feedback Control System
INTRODUCTION
Dr.-Ing. Erwin Sitompul
Chapter 1
1/6Erwin Sitompul Feedback Control System
Introduction
• Control is a series of actions directed for making a system variable adheres to a reference value (can be either constant or variable).
• The reference value when performing control is the desired output variable.
• Process, as it is used and understood by control engineers, means the component to be controlled.
Fundamental structures of control are classified based on the information used along the control process:
1. Open-loop control / Feedforward control2. Closed-loop control / Feedback control
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Process
Input
Performance
Measurement
DisturbanceReference
Measurement noise
1/8Erwin Sitompul Feedback Control System
The difference: In open-loop control, the system does not measure the
actual output and there is no correction to make the actual output to be conformed with the reference value.
In feedback control, the system includes a sensor to measure the actual output and uses its feedback to influence the control process.
Open-loop vs. Feedback Control
1/9Erwin Sitompul Feedback Control System
Examples
• The controller is constructed based on knowledge or experience.
• The process output is not used in control computation.
• The output is fed back for control computation.
Open-loop control Feedback control
Example: an electric toaster, a standard gas stove.
Example: automated filling up system, magic jar, etc.
1/10Erwin Sitompul Feedback Control System
Plus-Minus of Open-loop Control
+ Generally simpler than closed-loop control+ Does not require sensor to measure the output+ Does not, of itself, introduce stability problem
– Has lower performance to match the desired output compared to closed-loop control
1/11Erwin Sitompul Feedback Control System
Plus-Minus of Feedback Control
+ Process controlled by well designed feedback control can respond to unforeseen events, such as: disturbance, change of process due to aging, wear, etc.
+ Eliminates the need of human to adjust the control variable reduce human workload
+ Gives much better performance than what is possibly given by open loop control: ability to meet transient response objectives and steady-state error objectives
– More complex than open-loop control– May have steady-state error– Depends on the accuracy of the sensor– May have stability problem
Feedback Control System
DYNAMIC MODELS
Dr.-Ing. Erwin Sitompul
Chapter 2
1/13Erwin Sitompul Feedback Control System
Dynamic Models
A Simple System: Cruise Control ModelWrite the equations of motion for the speed and forward motion of the car shown below, assuming that the engine imparts a force u, and results the car velocity v, as shown.
Using the Laplace transform, find the transfer function between the input u and the output v.
u (Force)
x (Position)
v (Velocity)
1/14Erwin Sitompul Feedback Control System
Dynamic ModelsApplying the Newton’s Law for translational motion yields:
u bv ma u bx mx
b uv vm m
u bv mv
( )V s b m U m
( ) 1
( )
V s m
U s s b m
MATLAB (Matrix Laboratory) is the standard software used in control engineering:
In the end of this course, you are expected to be able to know how to use MATLAB for basic applications.
1/15Erwin Sitompul Feedback Control System
Dynamic ModelsWith the parameters:
1000 kg50 Ns/m500 N
mbu
( ) 1
( )
V s m
U s s b m
In MATLAB windows:
Response of the car velocity v to a step-shaped force u:
Time (sec)A
mp
litu
de
0 20 40 60 80 100 1200
1
2
3
4
5
6
7
8
9
10
1/16Erwin Sitompul Feedback Control System
Dynamic Models
A Two-Mass System: Suspension Modelm1 : mass of the wheelm2 : mass of the carx,y : displacements from equilibriumr : distance to road surface
s w 1( ) ( ) ( )k x y b x y k x r m x Equation for m1:
Equation for m2:
s 2( ) ( )k y x b y x m y
Rearranging:s w w
1 1 1 1
( ) ( )k k kb
x x y x y x rm m m m
s
2 2
( ) ( ) 0kb
y y x y xm m
1/17Erwin Sitompul Feedback Control System
Dynamic ModelsUsing the Laplace transform:
( ) ( )
( ) ( )
x t X sdx t sX s
dt
L
L
2
1
s w w
1 1 1
( ) ( ) ( )
( ) ( ) ( ) ( )
bs X s s X s Y s
mk k k
X s Y s X s R sm m m
2 s
2 2
( ) ( ) ( ) ( ) ( ) 0kb
s Y s s Y s X s Y s X sm m
to transfer from time domain to frequency domain yields:
1/18Erwin Sitompul Feedback Control System
w s
1 2
4 3 2s w w w s
1 2 1 2 1 1 2 1 2
( )
( )
k b ks
m m bY s
R s k k k b k kb b ks s s s
m m m m m mm mm
Dynamic ModelsEliminating X(s) yields a transfer function:
( )( )
( )
Y sF s
R s
outputtransfer function
input
1/19Erwin Sitompul Feedback Control System
Dynamic Models
Bridged Tee Circuit
1 o1 i1 1
1 2
0V VV V
sCVR R
o 12 o i
2
( ) 0V V
sC V VR
v1
Resistor Inductor Capacitor
dvi Cdt
v Ri div Ldt
( ) ( )V s R I s ( ) ( )V s sL I s ( ) ( )I s sC V s
1/20Erwin Sitompul Feedback Control System
Dynamic Models
1 o1 i 1 01 1
V VV V V
s
1 o o1 o
0( 1)
1
V V VV V s
s
o1 i
12
VV V
s s
oo i
11 2
VV s V
s s
o i2 3V s V
o
i
1
2 3
V
V s
RL Circuit
v1
Further calculation and eliminating V1,
Feedback Control System
DYNAMIC RESPONSE
Dr.-Ing. Erwin Sitompul
Chapter 3
1/22Erwin Sitompul Feedback Control System
Review of Laplace Transform
( )f t L
( )F s
( )G s 1L( )g t
Time domain Frequency domainProblem
Solution
easy operations
difficult operations
0
( ) ( ) ( ) stf t F s f t e dt
L
1 1( ) ( ) ( )
2
c
c
jst
j
F s f t F s e dsj
L
s j
1/23Erwin Sitompul Feedback Control System
Properties of Laplace Transform1. Superposition
1 2 1 2( ) ( ) ( ) ( )f t f t F s F s L
2. Time delay ( ) ( ) ( )sf t u t e F s L
3. Time scaling
1( )
sf at F
a a
L
4. Shift in Frequency ( ) ( ) ( )ate f t u t F s a L
5. Differentiation in Time
2
1 2 1
( ) ( ) (0 )
( ) ( ) (0 ) (0 )
( ) ( ) (0 ) (0 ) (0 )n n n n n
f t s F s f
f t s F s s f f
f t s F s s f s f f
LL
L
1/24Erwin Sitompul Feedback Control System
Properties of Laplace Transform6. Integration in Time
0
1( ) ( )
tf t dt F s
s L
7. Differentiation in Frequency
( )( )
dF st f t
ds L
8. Convolution
1 2 1 2
1 2 1 2
( ) ( ) ( ) ( )
( ) ( ) 2 ( ) ( )
F s F s f t f t
F s F s j f t f t
L
L
1 2 1 2
0
( ) ( ) ( ) ( )f t f t f f t d
1/25Erwin Sitompul Feedback Control System
t
( )t1
t
( )r t
1
1
unit impulse
unit step
unit ramp
Table of Laplace Transform
t
1( )t1
1/26Erwin Sitompul Feedback Control System
Example:Obtain the Laplace transform of
2( ) ( ) 2 1( ) 3 , 0.tf t t t e t
2( ) ( ) 2 1( ) 3 tF s t t e L L L
1 11 2 3
2s s
2 4
( 2)
s s
s s
2( ) 2 1( ) 3 tt t e L L L
Laplace Transform
1/27Erwin Sitompul Feedback Control System
Laplace Transform
Example:Find the Laplace transform of the function shown below.
t
( )g t4
0 1 2 3 4
( ) 4 1( 2) 4 1( 3)g t t t
2 3
( ) ( )
4 4s s
G s g t
e e
s s
L
2 34( )s se es
1/28Erwin Sitompul Feedback Control System
Inverse Laplace Transform
The steps are:
1. Decompose F(s) into simple terms using partial-fraction expansion.
2. Find the inverse of each term by using the table of Laplace transform.
Example:Find y(t) for
( 2)( 4)( ) .
( 1)( 3)
s sY s
s s s
31 2( )1 3
cc cY s
s s s
1 2 3( 1)( 3) ( 3) ( 1)
( 1)( 3)
c s s c s s c s s
s s s
1/29Erwin Sitompul Feedback Control System
Inverse Laplace Transform
1 2 3
1 2 3
1
14 3 6
3 8
c c cc c c
c
1
8,
3c 2
3,
2c 3
1
6c
8 1 3 1 1 1( )
3 2 1 6 3Y s
s s s
1
1 1 1
( ) ( )
8 1 3 1 1 1
3 2 1 6 3
y t Y s
s s s
L
L L L
38 3 11( ) 1( ) 1( )
3 2 6t tt e t e t
1 12 32 32
1 (4 3( )( )
(
3
1)( 3)
)s s cY s
s s
c c cc c c
s
2
( 1)( 3)
6 81s s
s s s
Comparing the coefficients
1/30Erwin Sitompul Feedback Control System
Initial and Final Value Theorem
0lim ( ) lim ( )
sty t s Y s
0lim ( ) lim ( )t sy t s Y s
0( ) lim ( ) lim ( )
t sy y t s Y s
Only applicable to stable system, i.e. a system with convergent step response
Example:Find the final value of the system corresponding to
2
3( 2)( )
( 2 10)
sY s
s s s
20
3( 2)( ) lim
( 2 10)s
sy s
s s s
3 2
0.610
1/31Erwin Sitompul Feedback Control System
Initial and Final Value Theorem
Example:Find the final value of the system corresponding to
0 0
3( ) lim ( ) lim ( ) lim
( 2)t s sy y t s Y s s
s s
3( )
( 2)Y s
s s
3
2
WRONGSince
3 3 2 3 2( )
( 2) 2Y s
s s s s
1 2( ) ( ) 3 2 1( ) 3 2 1( )ty t Y s t e t L NOT convergentNO limit value
1/32Erwin Sitompul Feedback Control System
Initial and Final Value Theorem
Example:Find the final value of
2
2( )
4Y s
s
20 0
2( ) lim ( ) lim ( ) lim
4t s sy y t s Y s s
s
0
WRONG
Since
2
2( ) ( ) sin 2
4Y s y t t
s
periodic signalNOT convergentNO limit value
1/33Erwin Sitompul Feedback Control System
Homework 12.63.4 (b)3.5 (c)3.6 (e)
Deadline: 10.05.2011, 7:30 am.